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I just want to know if $f $ is continuous on a compact interval, then does it follow that $\sqrt f$ is also continuous?

Stefan Hamcke
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    Yes, you just compose two continuous functions. (I presume that $f$ takes non-negative values only.) See this thread: http://math.stackexchange.com/questions/318614/composition-of-continuous-functions – Tomasz Kania Dec 13 '13 at 13:47

2 Answers2

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Suppose you have to prove that $\sqrt{f}$ is a continuous function (where $f$ is continuous). We go by the definition of continuity. Take any $\epsilon>0$. Then,

$$ \left|\sqrt{f(x)}-\sqrt{f(y)}\right|^2=f(x)+f(y)-2\sqrt{f(x)f(y)}. $$

Since $f$ should be non-negative,

$$ \begin{align} \left|\sqrt{f(x)}-\sqrt{f(y)}\right|^2&<f(x)+f(y)=\left|f(x)+f(y)\right|\\ \implies\quad\left|\sqrt{f(x)}-\sqrt{f(y)}\right|&<\sqrt{\left|f(x)+f(y)\right|}. \end{align} $$

Since $f$ is continuous we know that there exist a $\delta > 0$ such that,

$$ \left|f(x)+f(y)\right|<\epsilon^2\mbox{ whenever }|x-y|<\delta. $$

Hence,

$$\left|\sqrt{f(x)}-\sqrt{f(y)}\right|<\sqrt{\left|f(x)+f(y)\right|}<\epsilon\mbox{ whenever }|x-y|<\delta$$

and $\sqrt{f}$ is also continuous.

  • Could you please explain why continuity implies that $|f(x)+f(y)| \lt \varepsilon^2$? I thought it was $|f(x)-f(y)| \lt \varepsilon^2$ – S.C. May 08 '23 at 15:36
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Well, if $f\colon K\rightarrow \mathbb{R}_{+}$ is continuous, given that $\sqrt{\cdot} \ \colon \mathbb{R}_{+} \rightarrow \mathbb{R}_{+}$ is also continuous, then so is their composition: $\sqrt{f} \ \colon K \rightarrow \mathbb{R}_{+}$ (without any assumptions about K)

aaaaa
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  • Well you should at least assume $K$ is a topological space if you want the first statement to make sense. – JSchlather Dec 13 '13 at 14:08
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    True ... I was thinking of $K$ as a subset of the real line, and therefore, no special property (such as compactedness) was needed, but you're right. – aaaaa Dec 13 '13 at 14:45